已知在各項(xiàng)均不相等的等差數(shù)列{an}中,a1=1,且a1,a2,a5成等比數(shù)列,數(shù)列{bn}中,b1=log2(a2+1),bn+1=4bn+2n+1,n∈N*.
(Ⅰ)求{an}的通項(xiàng)公式及其前n項(xiàng)和Sn;
(Ⅱ)求證:{bn+2n}是等比數(shù)列,并求{bn}的通項(xiàng)公式;
(Ⅲ)設(shè)cn=akbk+2k,n=2k,k∈N*, 3×2k4bk-2k+1+2,n=2k-1,k∈N*,
求數(shù)列{cn}的前2n項(xiàng)的和T2n.
b
n
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1
=
4
b
n
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2
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1
{
b
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2
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a k b k + 2 k , n = 2 k , k ∈ N *, |
3 × 2 k 4 b k - 2 k + 1 + 2 , n = 2 k - 1 , k ∈ N *, |
【考點(diǎn)】錯(cuò)位相減法.
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發(fā)布:2024/6/27 10:35:59組卷:625引用:1難度:0.3
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